Tangents to the parabola

Quite a straightforward piece of maths I can't seem to get my head around here:

The tangent to the parabola at the point $(4a, 4a)$ is given by what equation?

Bearing in mind the parabola is $y^2 = 4ax$ and the general point $(at^2, 2at)$ lies on the parabola.

• Do you know about implicit differentiation? You need to find the derivative first – Sam Weatherhog Nov 4 '15 at 21:37
• @SamWeatherhog forgive me but what do I differentiate? – Pontius Pilate VII Nov 4 '15 at 21:39
• @PontiusPilateVII You differentiate both sides of $y^2=4ax$ with respect to $x$. The left-hand side is done via the chain rule to get $2yy'$. – Arthur Nov 4 '15 at 21:42

Do you know how to compute the tangents to $y=\frac{1}{4a}x^2$ by differentiation? If we take the point $(t,\frac{1}{4a}t^2)$, the tangent has equation $$y-\frac{1}{4a}t^2=\frac{1}{2a}t(x-t)$$ because the derivative of $x\mapsto \frac{1}{4a}x^2$ is $x\mapsto \frac{1}{2a}x$. For $t=4a$, we get $$y-\frac{1}{4a}16a^2=\frac{1}{2a}4a(x-4a)$$ that is $$y-4a=2x-8a$$ or $$y=2x-4a$$

Just swap $x$ with $y$ and the requested tangent will be $$x=2y-4a$$

Note that swapping $x$ with $y$ is an isometry, so it preserves tangents.

Let's make another example. Suppose you want to compute the normal to the parabola at the point $(a,-2a)$. You can do it in a very similar way, by computing the normal to the parabola $y=\frac{1}{4a}x^2$ at the point $(-2a,a)$. The tangent will be $$y-a=\frac{1}{2a}\cdot(-2a)(x+2a)$$ or $$y-a=-x-2a$$ that is $$y=-x-a$$ The normal is thus $$y=x-a$$ so the required normal is $$x=y-a$$ (by swapping back $x$ with $y$).

• @PontiusPilateVII If you swap $x$ with $y$ in your given equation, you get $x^2=4ay$, that is, $y=\dfrac{1}{4a}x^2$. – egreg Nov 4 '15 at 21:54
• @PontiusPilateVII Where's the difference? An isometry also preserves angles. Don't change the question so that it invalidates existing answers. – egreg Nov 4 '15 at 22:26
• Compute the tangent in the “swapped system”; the normal is easy to write; swap back. – egreg Nov 4 '15 at 22:29
• @PontiusPilateVII At the end, of course. – egreg Nov 4 '15 at 22:32
• @PontiusPilateVII I added the computation. – egreg Nov 4 '15 at 22:36